Problem: $f(x)=2x^4-10x^3-72x^2+100$. On which intervals is the graph of $f$ concave up? Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\infty,-\dfrac32\right)$ and $\left(4,\infty\right)$ (Choice B) B $\left(-\dfrac{3}{2},4\right)$ only (Choice C) C $\left(-3,4\right)$ only (Choice D) D $\left(-\infty,-3\right)$ and $\left(4,\infty\right)$
We can analyze the intervals where $f$ is concave up/down by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=12(2x+3)(x-4)$. $f''(x)=0$ for $x=-\dfrac{3}{2},4$. Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our points of interest are $x=-\dfrac{3}{2}$ and $x=4$. Our points of interest divide the number line into three intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $(-\infty, -\frac32)$ $\left(-\frac32,4\right)$ $\left(4,\infty\right)$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $\left(-\infty,-\dfrac{3}{2}\right)$ $x=-2$ $f''(-2)=72>0$ $f$ is concave up $\cup$ $\left(-\dfrac{3}{2},4\right)$ $x=0$ $f''(0)=-144<0$ $f$ is concave down $\cap$ $(4,\infty)$ $x=5$ $f''(5)=156>0$ $f$ is concave up $\cup$ In conclusion, the graph of $f$ is concave up over the intervals $\left(-\infty,-\dfrac32\right)$ and $\left(4,\infty\right)$.